From my understanding of limit points and interior points there is somewhat of an overlap and that a lot of the time interior points are also limit points.

For the reals, a neighborhood, $r>0$, around a point must contain only a single point of the set in question to determine if it's a limit point or not. However, an interior must be completely contained in the set in question, meaning it has a neighborhood that contains at least a point in the set, which also makes it a limit point.

For example: $[0,1)$ in the reals.

From what I understand the set $(0,1)$ in the set of interior points, the set $[0,1]$ in the set of all limit points, and the set $(0,1)$ is the set of all point which are both interior and limit points.

Is this correct or are interior points always not limit points for some reason?


You are right that interior points can be limit points.

Your example was a perfect one: The set $[0,1)$ has interior $(0,1)$, and limit points $[0,1]$. So actually all of the interior points here are also limit points.

Here is an example of an interior point that's not a limit point:

Let $X$ be any set and consider the discrete topology $\mathcal{T} = \mathcal{P}(X)$ on $X$. Then let $A$ be any non-empty subset of $X$. For any $a \in A$, $\{a \}$ is an open set containing $a$, and $a \in \{a \} \subseteq A$, so $a$ is an interior point for $A$. However, $a$ is not a limit point because we can find an open neighborhood of $a$ that doesn't contain any points from $A$ other than $a$ -- namely, $\{a \}$ is the open neighborhood that satisfies this. So $a$ is not a limit point of $A$. (By this same argument, we can show that $A$ has no limit points!)

  • $\begingroup$ You are saying {a} is an open set containing a, but it seems this is not an open set since there is no interval around a that is also in A. $\endgroup$ Sep 15 '19 at 11:30
  • 2
    $\begingroup$ @jeffery_the_wind You are speaking from the perspective of the Euclidean topology that we are all so used to. But in my answer, I mentioned that $\{a\}$ is open in the discrete topology, which is the topology that consists of all possible subsets of $X$. So every subset of $X$ is open, including $\{a\}$. So the question of which sets are open and which are closed really must be asked/framed in the context of a specific topology, since the answer changes if you change the topology. $\{a\}$ is not open in the Euclidean topology, but it is in the discrete topology. $\endgroup$
    – layman
    Sep 16 '19 at 3:23

A point $x$ of $A$ can be of one of two mutually exclusive types: a limit point of $A$ or an isolated point of $A$.

If the latter, it means that there exists some open $O$ in $X$ such that $\{x\} = O \cap A$. The negation of this is exactly that every open set $O$ that contains $x$ always intersects points of $A$ unequal to $x$ as well, and this means exactly that $x$ is a limit point of $A$.

E.g. $A = (0,1) \cup \{2,3\}$ (usual topology of the reals) has two isolated points $2$ and $3$ (which are not interior points of $A$), and the rest are limit points of $A$ as well as interior points. There are also limit points $0,1$ that are not in $A$ (showing $A$ is not closed).

So if $A$ has no isolated point, all of the points of $A$ (in particular all its interior points) are limit points of $A$. So often there will quite an overlap between interior and limit points.


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